A 6.0-μF air capacitor is connected across a 100-V battery. After the battery fully charges the capacitor, the capacitor is immersed in transformer oil (dielectric constant = 4.5). How much additional charge flows from the battery, which remained connected during the process?

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Vespertilio Vespertilio · Answer 1 · 2019-08-04T17:04:13+02:00

Answer:

21 x 10^-4 C

Explanation:

C = 6 micro Farad = 6 x 10^-6 F

V = 100 V

Charge, q = C V = 6 x 10^-6 x 100 = 6 x 10^-4 C

As the dielectric is inserted, the capacitance is increased.

C' = K C = 4.5 x 6 x 10^-6 = 27 x 10^-6 F

As the battery remains connected, so the voltage across the plates remains same and charge becomes q'.

V = q' / C'

q' = C' V = 27 x 10^-6 x 100 = 27 x 10^-4 C

Amount of additional charge = q' - q = 27 x 10^-4 - 6 x 10^-4 = 21 x 10^-4 C